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Ludmilka [50]
3 years ago
11

Explain how the gas laws apply to the act of breathing. Describe the changes that occur in each step of the process in terms of

the pressure, volume, temperature, and moles of gas present for each step, identify which variables are constant, and which are changing name the gas law or gas laws that apply to the process create and solve a sample calculation using realistic values of pressure, temperature, amount of gas, and volume. Please HELP!
Physics
1 answer:
MrRa [10]3 years ago
4 0
<span>When you in inhale, you move your diaphragm down increasing the volume of the cavity in your ribs (its called pleural cavity). This lowers the pressure inside your lungs. When the pressure in your lungs is less than the atmospheric pressure, air draws inwards. Ideal gas law: PV=nRT -> assume nRT are constant for our purposes. then we allow one system to be equal to the same system at different conditions, thus: P1V1=P2V2 (Boyles Law) assume P1=1 atm and V1=1 L (keeping the numbers simple to avoid confusion). when the V is increased (for example say doubled) V2=2 L, then: (1 atm)*(1 L)=P2*(2 L) -> P2=0.5 atm The pressure was decreased. Apply the same logic for exhalation.</span>
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stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

-0.001\times 2vdv=dt

vdv=-\frac{dt}{0.002}

at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

3 0
3 years ago
to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on
Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

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So, μ = mg/kx

8 0
3 years ago
a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0
3 years ago
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put i
Margaret [11]

Answer: 1.176×10^-3 s

Explanation: The time constant formulae for an RC circuit is given below as

t =RC

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t = 56×10^-6 × 21

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4 0
3 years ago
Read 2 more answers
Anybody know these ?
pashok25 [27]

Answer:

1. a) 72 N.

2. a) 2 m/s².

Explanation:

Given the following data;

1. Mass = 90kg

Acceleration = 0.8 m/s²

To find the force;

Force = mass * acceleration

Force = 90 * 0.8

Force = 72 Newton.

2. Mass = 50kg

Force = 100N

To find the magnitude of acceleration;

Acceleration = force/mass

Acceleration = 100/50

Acceleration = 2 m/s²

5 0
3 years ago
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